What is the remainder when 2 + 4 + 6 + ... + 100 + 102 + 104 is divided by 7?
+37153 Arithmetic sequence with 52 terms
sum = 52 ( 2+104)/2 =2756 2756 mod 7 =5
+15 There are 52 terms in the sequence. If you divide each of the numbers in 2: 2,4,6,...,102,104 = 1,2,3,...,51,52.
2+104 = 4+102 = 6+100 which means ((2+104)x52)/2 equals the sum of the sequence, which is 2756, and if you divide that by 7 you get 393 r5.
So, the answer is 5.
